I am just playing a little with this question and looking at Gino's answer.
I am not familiar with this topic.
This quiet froum is great I have time to go back and look at some of the old questions and answers 
I need to find the asymptotes of this hyperbola (x+2)^2/121-(y-5)^2/81=1
\(\frac{(x+2)^2}{121}-\frac{(y-5)^2}{81}=1\\ centre\;\;(-2,5)\\\)
The square roots of the denominators of the x-term and the y-term give the slope of the lines. Since slope is y-value over x-value, the slopes will be 9/11 and -9/1
\(gradients = \frac{\pm \sqrt{y\; denominator}}{\sqrt{x\; denominator}}=\frac{\pm \sqrt{81}}{\sqrt{121}}=\pm \frac{ 9}{11}\)
Why?
The equations will be y - 5 = ± (9/11)(x + 2).
When graphing, you draw the box; the asymptotes are the lines created by the opposite corners of the box. ???
Mmm Interesting....... More thought needed .......
+118729 
